ÌâÄ¿ÄÚÈÝ

16£®ÏòNaAlO2ÈÜÒºÖÐÖðµÎ¼ÓÈëÏ¡ÑÎËᣬÉú³É³ÁµíµÄÁ¿ÏÈÖð½¥Ôö¶àºóÓÖÖð½¥¼õÉÙÖÁÇ¡ºÃÏûʧ£®ÏÂÁÐÓйØÐðÊöÕýÈ·µÄÊÇ£¨¡¡¡¡£©
A£®µ±³ÁµíÇ¡ºÃÏûʧʱc£¨Cl-£©=3c£¨Al3+£©
B£®Èô½«·´Ó¦ºóµÄÈÜÒºÓëÔ­NaAlO2ÈÜÒº»ìºÏ£¬ÔòÇ¡ºÃΪÖÐÐÔ
C£®NaAlO2ÈÜÒºÖÐÀë×ÓŨ¶È´óС˳ÐòΪ£ºc£¨AlO2-£©£¾c£¨Na+£©£¾c£¨OH-£©£¾c£¨H+£©
D£®µ±c£¨Na+£©=c£¨Cl-£©Ê±³ÁµíÁ¿´ïµ½×î´óÖµ

·ÖÎö ÏòNaAlO2ÈÜÒºÖÐÖðµÎ¼ÓÈëÏ¡ÑÎËᣬÉú³É³ÁµíµÄÁ¿ÏÈÖð½¥Ôö¶àºóÓÖÖð½¥¼õÉÙÖÁÇ¡ºÃÏûʧ£¬ÔòÆ«ÂÁËáÄÆºÍÑÎËá·´Ó¦ÏÈÉú³ÉÇâÑõ»¯ÂÁ³Áµí£¬ºóÇâÑõ»¯ÂÁºÍÑÎËá·´Ó¦Éú³É¿ÉÈÜÐÔµÄÂÈ»¯ÂÁ£¬·¢ÉúµÄ·½³ÌʽΪ£ºNaAlO2+HCl+H2O=Al£¨OH£©3¡ý+NaCl£¬Al£¨OH£©3¡ý+3HCl=AlCl3+3H2O£¬×Ü·´Ó¦Îª£ºNaAlO2+4HCl=AlCl3+2H2O+NaCl£¬¾Ý´Ë·ÖÎö½â´ð£®

½â´ð ½â£ºA¡¢µ±³ÁµíÇ¡ºÃÏûʧʱ£¬¼´·¢Éú×Ü·´Ó¦Îª£ºNaAlO2+4HCl=AlCl3+2H2O+NaCl£¬ËùÒÔc£¨Cl-£©=4c£¨Al3+£©£¬¹ÊA´íÎó£»
B¡¢¸ù¾ÝÒÔÉÏ·ÖÎö£¬·´Ó¦ºóµÄÈÜҺΪµÈÎïÖʵÄÁ¿µÄAlCl3ºÍNaCl£¬ÓëÔ­NaAlO2ÈÜÒº»ìºÏ£¬AlCl3ºÍNaAlO2ÈÜÒº·¢Éú˫ˮ½âÉú³ÉÇâÑõ»¯ÂÁ³Áµí£¬·½³ÌʽΪ£ºAlCl3+3NaAlO2+6H2O=4Al£¨OH£©3¡ý+3NaCl£¬ÔòAlCl3¹ýÁ¿Ë®½âÏÔËáÐÔ£¬ËùÒÔ·´Ó¦ºóÈÜÒºÈÔȻΪËáÐÔ£¬¹ÊB´íÎó£»
C¡¢NaAlO2ÈÜÒºAlO2-·¢ÉúË®½âÏÔ¼îÐÔ£¬ËùÒÔÀë×ÓŨ¶È´óС˳ÐòΪ£ºc£¨Na+£©£¾c£¨AlO2-£©£¾c£¨OH-£©£¾c£¨H+£©£¬¹ÊC´íÎó£»
D¡¢¸ù¾ÝÒÔÉÏ·ÖÎö£¬³ÁµíÁ¿´ïµ½×î´óֵʱ·¢Éú·´Ó¦ÎªNaAlO2+HCl+H2O=Al£¨OH£©3¡ý+NaCl£¬ËùÒÔµ±c£¨Na+£©=c£¨Cl-£©Ê±³ÁµíÁ¿´ïµ½×î´óÖµ£¬¹ÊDÕýÈ·£»
¹ÊÑ¡£ºD£®

µãÆÀ ±¾Ì⿼²éÁËÂÁ¼°Æä»¯ºÏÎïµÄÐÔÖʵķÖÎöÓ¦Óã¬ÕÆÎÕÎïÖÊÐÔÖÊÊǽâÌâ¹Ø¼ü£¬ÌâÄ¿ÄѶÈÖеȣ®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø